If the maximum velocity and acceleration of a particle executing SHM are equal in magnitude the time period will be

Options

A. 1.57 s
B. 3.14 s
C. 6.28 s
D. 12.56 s

Explanation

To solve the problem, we need to establish the relationship between maximum velocity, maximum acceleration, and the time period of a particle executing Simple Harmonic Motion (SHM).

### Step-by-Step Solution:

1. **Understanding SHM Parameters**:
- In SHM, the maximum velocity \( V_{\text{max}} \) is given by the formula:
\[ V_{\text{max}} = A \cdot \omega \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.

2. **Maximum Acceleration**:
- The maximum acceleration \( A_{\text{max}} \) in SHM is given by:
\[ A_{\text{max}} = A \cdot \omega^2 \]

3. **Setting the Condition**:
- According to the problem, the maximum velocity and maximum acceleration are equal in magnitude:
\[ V_{\text{max}} = A_{\text{max}} \]
- Substituting the formulas from steps 1 and 2:
\[ A \cdot \omega = A \cdot \omega^2 \]

4. **Simplifying the Equation**:
- Since \( A \) (amplitude) is not zero, we can divide both sides of the equation by \( A \):
\[ \omega = \omega^2 \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ \omega^2 - \omega = 0 \]
- Factoring out \( \omega \):
\[ \omega(\omega - 1) = 0 \]

6. **Finding the Values of \( \omega \)**:
- This gives us two solutions:
- \( \omega = 0 \) (not applicable since it would imply no motion)
- \( \omega = 1 \)

7. **Calculating the Time Period**:
- The time period \( T \) is related to the angular frequency \( \omega \) by the formula:
\[ T = \frac{2\pi}{\omega} \]
- Substituting \( \omega = 1 \):
\[ T = \frac{2\pi}{1} = 2\pi \approx 6.28 \text{ seconds} \]

### Final Answer:
The time period \( T \) of the particle executing SHM when the maximum velocity and acceleration are equal in magnitude is approximately \( 6.28 \) seconds.

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